What is the integral of csc(x)sec(x) ?

The integral of csc(x)sec(x) is 1/2 ln(tan^2(x))+C

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integral of csc(x)sec(x)

\int csc (x) sec (x) d x

\frac{1}{2} ln (tan^{2} (x)) + C

Solution steps

\int csc (x) sec (x) d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 1} d u

= \int \frac{u}{u ^{2} - 1} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln sec^{2} (x) - 1

Simplify

\frac{1}{2} ln sec^{2} (x) - 1 : \frac{1}{2} ln (tan^{2} (x))

= \frac{1}{2} ln (tan^{2} (x))

Add a constant to the solution

= \frac{1}{2} ln (tan^{2} (x)) + C

f (x) = 6 x

f (x) = 3 e^{(2 x^{2})}

(y - 9)^{2} = y^{^{'}}

\frac{t}{t ^{2} + 1}

\int 5 e^{7 x} d x

What is the integral of csc(x)sec(x) ?
The integral of csc(x)sec(x) is 1/2 ln(tan^2(x))+C